You have found the following ages (in years) of all 5 seals at your local zoo: $ 28,\enspace 12,\enspace 10,\enspace 1,\enspace 4$ What is the average age of the seals at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{28 + 12 + 10 + 1 + 4}{{5}} = {11\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $28$ years $17$ years $289$ years $^2$ $12$ years $1$ year $1$ year $^2$ $10$ years $-1$ years $1$ year $^2$ $1$ year $-10$ years $100$ years $^2$ $4$ years $-7$ years $49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{289} + {1} + {1} + {100} + {49}} {{5}} $ $ {\sigma^2} = \dfrac{{440}}{{5}} = {88\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{88\text{ years}^2}} = {9.4\text{ years}} $ The average seal at the zoo is 11 years old. There is a standard deviation of 9.4 years.